To solve $\sin \left( - \frac{5\pi}{12} \right)$, we can use the identity for the sine of a negative angle:

$\sin(-\theta) = -\sin(\theta)$

This means:

$\sin \left( - \frac{5\pi}{12} \right) = - \sin \left( \frac{5\pi}{12} \right)$

Next, we can break $\frac{5\pi}{12}$ into known angles that we can work with. Notice that:

$\frac{5\pi}{12} = \frac{\pi}{4} + \frac{\pi}{6}$

We will now use the angle addition formula for sine:

$\sin(a + b) = \sin(a) \cos(b) + \cos(a) \sin(b)$

Substituting $a = \frac{\pi}{4}$ and $b = \frac{\pi}{6}$, we get:

$\sin \left( \frac{5\pi}{12} \right) = \sin \left( \frac{\pi}{4} + \frac{\pi}{6} \right) = \sin \left( \frac{\pi}{4} \right) \cos \left( \frac{\pi}{6} \right) + \cos \left( \frac{\pi}{4} \right) \sin \left( \frac{\pi}{6} \right)$

Using known values:

$\sin \left( \frac{\pi}{4} \right) = \cos \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2}$ $\sin \left( \frac{\pi}{6} \right) = \frac{1}{2}, \quad \cos \left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2}$

Substitute these into the formula:

$\sin \left( \frac{5\pi}{12} \right) = \left( \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} \right) + \left( \frac{\sqrt{2}}{2} \times \frac{1}{2} \right)$

Simplify each term:

$\sin \left( \frac{5\pi}{12} \right) = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4}$

So:

$\sin \left( \frac{5\pi}{12} \right) = \frac{\sqrt{6} + \sqrt{2}}{4}$

Finally, remembering the negative sign:

$\sin \left( -\frac{5\pi}{12} \right) = - \frac{\sqrt{6} + \sqrt{2}}{4}$

Final Answer:

$\sin \left( - \frac{5\pi}{12} \right) = - \frac{\sqrt{6} + \sqrt{2}}{4}$

Would you like more details on the solution or have any questions?

Here are some related questions you might find interesting:

1. How do you derive the angle addition formula for sine?
2. What other trigonometric identities can simplify expressions with negative angles?
3. How do you evaluate sine and cosine for common angles like $\frac{\pi}{6}$, $\frac{\pi}{4}$, and $\frac{\pi}{3}$?
4. What is the exact value of $\sin \left( \frac{\pi}{12} \right)$?
5. How can you use trigonometric identities to simplify sums of angles in radians?

Tip: Remember that recognizing common angle relationships is key to simplifying complex trigonometric expressions quickly!